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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Jul 2007 19:51:45 IST
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evaluate [ ] [ ] x 2 log (1-x 2) dx....hence prove that 1/1.5 + 1/2.7 + 1/3.9 + ............... = 8/9 - 2/3log2 rates assured....
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30 Jul 2007 14:20:31 IST
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?????????????????????????
reply plz
?????????????????????????
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30 Jul 2007 19:57:37 IST
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heeeeeeeeelllllllllllllllllllllllllllllppppppppppppppppppppppppppppppppppppppp
plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
kahan ho sab......
no nickels to ask.....helpppp :( :( :( :( :(
himanshu bhai karo solve iseeeeee
nebody tryyyyyyyyy
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hiee..."kahan ho bhai " hmmm...well here's the sol.... we all know log(1-x) = - { x + x2/2 + x3/3 + ...............infinity} see now x can be nethng , i mean x is wat we assume , so we can put x2 also in place of x putting x2 in place of x , identity becomes log (1-x2) = - {x2 + x4/2 + x6/3.....infinity} multiplying both sides by x2 , we get x2 log(1-x2) = -{x4 + x6/2 + x8/3 + ..........infinity} now integrating both the sides... [ ][ ] x2 (log (1-x2))dx = - {x ^5 /5 + x 7/14 + x9/27 + ......infinty} it can be written as x 2(log (1-x 2) dx = - {x5/1.5 + x7/2.7 + x9/3.9 + ........infinity} + c to find c , put x = 0 0 = 0 + c , so c = 0 x2(log (1-x2))dx = - {x5/1.5 + x7/2.7 + x9/3.9 +.........infinity} integrating by parts and taking limits 0 to 1 [(x3/3)log(1-x2)]10 - integ. (frm 0 to 1) x3 / 3 (-2x/1-x2) =>[(x3/3)log(1-x2)]10 + 2/3 [ -x3/3 - x + 1/2log(1+x/1-x)]10 log (1-x2) = log (1+x) + log (1-x) and log (1+x/1-x) = log (1+x) - log (1-x) (note this step now) =>1/2 log2 + 1/3log2 -2/3 -2/9 + lim.(x->1) (x3/3-1/3) log(1-x) (because limx->1) x3-1log(1-x) = 0 so it becomes 2/3log2 - 8/9 so LHS = RHS hence proved..... phew...
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I always like to walk in rain as no one can see me crying there :(
frnds are like diamonds , if u hit them , they don't break but they slip frm ur hands
-----It is better to be hated for what you are than to be loved for what you are not.----
*****wen love and skill work together--expect a masterpiece*****
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30 Jul 2007 21:39:29 IST
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thanx a tonnn........but cannot rate  .....neways aise qws nahi aate kya exam me????????
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30 Jul 2007 22:54:37 IST
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1 more....
integral..(dx / tanx + cotx + secx + cosecx)
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31 Jul 2007 00:22:09 IST
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arey bhai , isme kya hai????? its simple yaar...chal fir bi here's the sol.... [ ][ ] (dx / tanx + cotx + secx + cosecx) changing all terms into sin x and cos x => dx / (sinx/cosx) + (cosx/sinx) + (1/cosx) +(1/sinx) taking lcm => sinxcosx (dx) / sin 2 x + cos 2x + sinx + cosx sin2x + cos2x = 1 => sinxcosxdx / 1 + sinx + cosx taking cosx common frm den. => sinx dx / secx + tanx + 1 ............(u can understand this step) now multiplying and dividing by (1 + tanx - secx)............note this step => sinx (1+tanx - secx)dx / (1+tanx) 2 - sec 2x => sinx (1+tanx -secx)/2tanx opening tanx in den. , sin x gets cancelled => 1/2 cosx(1+tanx-secx)dx => 1/2 (cosx + sinx - 1) dx..... = 1/2 (sinx - cosx - x) + c......... its the answer cheers bhai..... aur koi qs hai????????????
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I always like to walk in rain as no one can see me crying there :(
frnds are like diamonds , if u hit them , they don't break but they slip frm ur hands
-----It is better to be hated for what you are than to be loved for what you are not.----
*****wen love and skill work together--expect a masterpiece*****
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31 Jul 2007 13:01:55 IST
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thank uuuuuuu :) :) :) :) :)
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2 Aug 2007 18:52:34 IST
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me too wanted the answers....thanx a lotttttt
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